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After watching this video lesson, you will be able to find the right answer for a trigonometric equation with a restricted domain. Learn what it means to have a restricted domain and what you need to do to find your answer.

Trigonometric Equations

In this video lesson, we will look at solving trigonometric equations. These are our equations involving the trigonometric functions. So, our variable will be part of the trig function's argument. Do you remember your trig functions? We have our basic ones of sine, cosine, and tangent. We also have cosecant, secant, and cotangent.

So, our trigonometric equations will have at least one of these functions in the equation, such as f(x) = sin (x) + 2. It could even have a combination of trig functions such as f(x) = cos (x) + cot (x). Notice how our variable is part of the argument in both of these equations? This is what makes these equations trigonometric equations.

Restricted Domains

In this video lesson, we do have one additional criteria that we are adding onto our trigonometric equations. And this is called a restricted domain. What does this mean? It means that we are limiting the possible values that our variable can be. Because we are working with trigonometric equations, our variable will be in radian units. So, you will most likely see a pi in the restricted domain.

For example, you might see f(x) = sin (x) + 2; 0 < x < pi.

This tells you that if you are solving this function, then you are looking for the solution that lies between 0 and pi. Remember that our trigonometric equations are all periodic, which means that the solutions repeat every so often. For sine and cosine, the standard period is 2pi. For tangent, it is pi. Remember that your calculator only gives you the primary solution and you might have to do some additional calculations to find the right answer that is within the restricted domain. Sometimes, if our domain is too restrictive, though, we may not have a solution at all. Let's take a look at a couple of examples.

Example 1

Solve f(x) = 5 cos (x); 0 <= x <= 2pi.

This problem is telling us to find the solutions of f(x) = 5 cos (x). This we can easily do. But wait, it also tells us that our answers are limited to between 0 and 2pi inclusive. So, this means that when we find our answers, we have to make sure they are within this range. Let's see what comes of this.

Solving for x means we set our equation to 0. We get 5 cos (x) = 0. To solve for x, we divide by 5 on both sides and then take an arccosine. We get x = cos^-1 (0). What does this equal? Our calculator tells us that it equals 90 degrees or pi/2 radians. Are we done though? Have we found all the answers? Let's graph out our function to see:

Graph of example problem

I have marked our restricted domain by labeling the x = 0 point and the x = 2pi point. We just look between these two points to see how many times our graph crosses the x axis. We see that it crosses the x axis twice between these two points. That means that we have a total of two answers. We found our first one, pi/2. So, what is the second one?

I have already labeled it on this graph. But if this point isn't labeled for you, then you can refer to your unit circle to help you find it. Looking at our unit circle, we see that another point where cos (x) is equal to 0 is when x = 3pi/2:

Unit Circle

So, our two answers are pi/2 and 3pi/2. Both of these answers are within the limits of our domain, so both are valid answers. And now we are done.

Example 2

Let's take a look at one more.

Solve f(x) = sin (x) - 1; 2pi < x < 3pi.

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This time, our problem is telling us to solve the equation f(x) = sin (x) - 1. We can do that by setting this equation equal to 0.

sin (x) - 1 = 0.

But wait, this time our domain is limited to between 3pi and 4pi non-inclusive. Hmm, so our answer must lie between 2pi and 3pi. Okay. Let's solve our equation to see what we get first. Adding 1 to both sides, we get sin (x) = 1. Taking the arcsine of both sides, we get x = sin^-1 (1). The arcsine of 1 gives us 90 degrees or pi/2 radians. Hmm. This is not within our domain.

So, what do we do? We remember that our sine function has a period of 2pi, so that means our solutions repeat every 2pi. So, we can add 2pi to pi/2. We get 5pi/2. Aha, this is between 2pi and 3pi. So, 5pi/2 is one of our answers. Do we have more, though? Let's graph this equation to see if we have more solutions within our domain.

Graph of example problem

I have marked our domain using blue vertical lines. So, we are looking for a solution between those lines. How many times does our graph cross or touch the x axis between these lines? Only once, so that means we only have one solution. We already have that one solution: 5pi/2. So, we are done!

Lesson Summary

Let's review what we've learned now. We learned that trigonometric equations are the equations that include trigonometric functions. A restricted domain means that we are limiting the possible values that our variable can be.

If your problem has a restricted domain, it will show it right next to the problem. When we have a problem with a restricted domain, it means that we need to find solutions that are within the restricted domain. We remember that our trig functions are periodic. Our sine and cosine functions have a standard period of 2pi, and our tangent function has a standard period of pi. If a function is periodic, it means that solutions repeat every so often. When solving, our calculators will only give the primary answer. If this answer is outside the restricted domain, then we need to make an additional calculation based on the period of our equation to find the right answer within our restricted domain.

Learning Outcomes

When the lesson ends, test your knowledge by doing the following:

Recognize trigonometric equations and restricted domain

Solve trigonometric problems with restricted domains

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